Multiply the following complex numbers: $({2-4i}) \cdot ({3})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({2-4i}) \cdot ({3}) = $ $ ({2} \cdot {3}) + ({2} \cdot {0}i) + ({-4}i \cdot {3}) + ({-4}i \cdot {0}i) $ Then simplify the terms: $ (6) + (0i) + (-12i) + (0 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 6 + (0 - 12)i + 0i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 6 + (0 - 12)i - 0 $ The result is simplified: $ (6 - 0) + (-12i) = 6-12i $